# Equilateral triangles in the complex plane

I came across an exercise in Ahlfors’ Complex Analysis the other day that got me thinking. The exercise asked to prove that the complex numbers $a$, $b$, and $c$ form the vertices of an equilateral triangle if and only if $a^2+b^2+c^2=ab+bc+ca.$ It struck me as quite a nice, simple, and symmetric condition. My first instinct, in going about proving this, was to see if the condition was translation invariant, so that one of the points can be moved to the origin. Indeed, if we subtract a constant $z$ from each of $a,b,c$ the equation becomes
$(a-z)^2+(b-z)^2+(c-z)^2=(a-z)(b-z)+(b-z)(c-z)+(c-z)(a-z),$
which simplifies to the original equation after expanding each term. So, we can assume without loss of generality that $a=0$, and we wish to show that $0$, $b$, and $c$ form the vertices of an equilateral triangle if and only if $b^2+c^2=bc$.

To finish up, since we can multiply $b$ and $c$ by a constant scaling factor without changing the condition, we can further assume that $b=1$. So then $1+c^2=c$, implying that $c=\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$. So indeed, the three points form an equilateral triangle. QED.

This proof works, but is somehow deeply unsatisfying. I wanted to find a more “symmetric” proof, that didn’t involve moving one of the points to an origin and another to an axis. Such a coordinate-free condition should have a coordinate-free proof.

Another way to approach it is to first manipulate the condition a bit:
\begin{eqnarray*}
a^2+b^2+c^2-ab-bc-ca&=& 0 \\
2a^2+2b^2+2c^2-2ab-2bc-2ca &=& 0 \\
(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2) &=& 0 \\
(a-b)^2+(b-c)^2+(c-a)^2=0
\end{eqnarray*}

So, we have re-written the equation as a sum of three squares that equals $0$. (Too bad we’re looking for complex and not real solutions!)

Setting $x=a-b$, $y=b-c$, and $z=c-a$, the condition now becomes $x^2+y^2+z^2=0$, along with the additional equation $x+y+z=0$. Notice that $x$, $y$, and $z$ are, in some sense, the sides of the triangle $abc$ as vectors. So, we are trying to show that if $x^2+y^2+z^2=0$ and $x+y+z=0$, then $x$, $y$, and $z$ all have the same distance to $0$ and are spaced $120^\circ$ apart from each other around the origin. This is starting to sound more tractable.

In fact, upon closer inspection, we are simply studying the intersection of two curves in the complex projective plane. The points in this plane are equivalence classes of triples $(x:y:z)$, not all zero, up to scaling by a complex number (which is dilation and rotation, geometrically). So, we can think of the equation $x^2+y^2+z^2=0$ as a quadratic curve in this plane, and the equation $x+y+z=0$ as a line, and we simply want to know where these curves intersect.

Intuitively, a quadratic curve (such as a circle or a parabola) and a line should intersect in at most 2 points. Indeed, Bezout’s theorem tells us that two complex projective curves of degrees $d$ and $e$ intersect in at most $de$ points. Here, $d$ and $e$ are $2$ and $1$, so we have at most $2$ solutions $(x:y:z)$ to the equations $x^2+y^2+z^2=0$ and $x+y+z=0$.

But we already know two solutions: first, any $x,y,z$ that are equal in norm and spaced at $120^\circ$ angles around $0$ will satisfy $x+y+z=0$, since they form an equilateral triangle with its centroid at the origin. Since squaring them will double their angles, $x^2$, $y^2$, and $z^2$ are at angles of $240^\circ$ from each other, which is the same as $120^\circ$ in the opposite direction! So, $x^2+y^2+z^2=0$ as well in this case.

Moreover, such $x$, $y$, and $z$ are unique up to dilation, rotation, and reflection of the equilateral triangle (say, interchanging $x$ and $y$). But scaling or rotating $x$, $y$, and $z$ by a complex constant does not change the point $(x:y:z)$, and so the only other distinct point is formed by reflecting the triangle. (Thanks to Bryan Gillespie for this helpful geometric insight!)

So we already know two distinct solutions, and by Bezout’s theorem there cannot be any others. It follows that the original triangle $abc$ must be equilateral.

Somehow more satisfactorily, QED.

## 6 thoughts on “Equilateral triangles in the complex plane”

1. Steven Karp on said:

Hi Maria,

That’s a very enlightening solution. You might also be interested in this approach. Letting $u := frac{a+b+c}{3}$ be the centroid of the triangle, we see that $a^2 + b^2 + c^2 = ab + bc + ca$ is equivalent to $3u^2 = a^2 + b^2 + c^2$. Using the identity $(x-a)(x-b)(x-c) = (x-u)^3 + frac{3u^2 – a^2 – b^2 – c^2}{2}x + u^3 – abc$ in $mathbb{C}[x]$, we see that this is in turn equivalent to $(a-u)^3 = (b-u)^3 = (c-u)^3$. This equation holds if the triangle is equilateral, and conversely, if this equation holds then $u$ is the circumcenter of the triangle, so the triangle is equilateral.

Do you see a way to generalize this result?

Happy studying!

2. Steven Karp on said:

Hi Maria,

I came across what I think is ‘the’ proof of this theorem. Note that two nondegenerate polygons in the plane are similar iff there exists a spiral similarity which takes one to a translation of the other. (Here similarities preserve orientation, i.e. reflections are not allowed.) In the complex plane, a spiral similarity is precisely multiplication by a nonzero complex number. Hence if $p_1, ldots, p_n$ and $q_1, ldots, q_n$ are the (ordered) vertices of these polygons, then the polygons are similar iff the vectors $mathbb{1}$, $p$, and $q$ are linearly dependent. Also note that $triangle ABC$ is equilateral iff $triangle ABCsimtriangle BCA$. Hence $a,b,c$ are the vertices of an equilateral triangle iff $detleft(begin{bmatrix}1 & a & b \ 1 & b & c \ 1 & c & aend{bmatrix}right) = 0$.

3. Chuck Ullery on said:

Thanks for the wonderful post about the equilateral triangles in the complex plane. I am a soon to be retired musician, and have decided to go through Ahlfors, but I got stuck on that problem. Beautiful! My only math experience is as an undergrad in the 60s, a brief try at grad school before being drafted, and as the homework helper for my daughter several years ago. She is now a recent PhD in algebraic geometry, and I am a very proud dad! I REALLY appreciate what you are doing with this blog. I’ll be back. Thanks again!!

• Maria Gillespie on said:

Thank you for your comment and interest – it’s great to hear from the visitors to my blog that I haven’t met before. Congrats to your daughter and keep up the good work on Ahlfors! 🙂

4. Ed Fahy on said:

Maria, Nice blog!
Excuse my notation; I don’t know how to do your nice Latex reps.
a^2+b^2+c^2 = ab + bc + ca, iff the same is true for a’, b’ and c’ with:
a’=exp(i al) (a-zo); b’ = exp(i al) (b-zo); c’ = exp(i al) (c-zo)
Like you explain earlier, the later is just a translation a, b and c by an amount zo and then a rotation about the origin of al. This translation and rotation can be done to make number a lie on the x-axis and because its an equilateral triangle, numbers b and c be equal to number a rotated through 120 degrees in the plus and minus direction, respectively.
[Let zo be the centroid of a, b and c: zo = (a+b+c)/3; and al = -arg(a-zo).]
Then, original problem on equilateral triangles is equivalent to:
a^2+b^2+c^2 = ab + bc + ca iff zo and al can be found such that
a’, b’= a’ exp(i 2pi/3), c’ = a’ exp(-i 2pi/3) (this is the general equilateral triangle with centroid at the origin and number a on the x-axis)
For this choice of a’, b’ and c’:
a’ ^2 +b’^2 +c’^2 = a’b’ + b’ c’ +c’ a’ = a’^2 [1+exp(+i 2 pi/3) + exp(-i 2 pi/3)]

5. Ed Fahy on said:

Interesting this relationship doesn’t extend to squares, 5-sided figures, etc. However, a similar type of relationship applies if the figure’s centroid is at the origin of the complex plane, but otherwise, doesn’t work.